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Question 147 of 148

Q.If lim⁡θ→0(1−cos⁡mθ1−cos⁡nθ)=1\displaystyle\lim_{\theta\to0}\left(\dfrac{1-\cos m\theta}{1-\cos n\theta}\right)=1, then prove that m=±nm=\pm n

Puducherry TnboardTamil Nadu HSC (DGE) Board 2026Subjective· 3mImportance★★★★★
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Rewrites 1−cos⁡x1-\cos x using the half-angle identity, isolates the standard sin⁡ϕϕ→1\frac{\sin\phi}{\phi}\to1 limit form, evaluates the limit as m2/n2m^2/n^2, and sets it equal to 11.

  1. Use the identity 1−cos⁡x=2sin⁡2(x2)1-\cos x=2\sin^2\left(\dfrac x2\right) for both numerator and denominator: 1−cos⁡mθ1−cos⁡nθ=2sin⁡2(mθ2)2sin⁡2(nθ2)=sin⁡2(mθ2)sin⁡2(nθ2)(ne0)\dfrac{1-\cos m\theta}{1-\cos n\theta}=\dfrac{2\sin^2\left(\frac{m\theta}2\right)}{2\sin^2\left(\frac{n\theta}2\right)}=\dfrac{\sin^2\left(\frac{m\theta}2\right)}{\sin^2\left(\frac{n\theta}2\right)}\quad(n e0)
  2. Multiply and divide to expose the standard limit form: sin⁡2(mθ2)sin⁡2(nθ2)=(sin⁡(mθ2)mθ2)2(nθ2sin⁡(nθ2))2⋅m2n2\dfrac{\sin^2\left(\frac{m\theta}2\right)}{\sin^2\left(\frac{n\theta}2\right)}=\left(\dfrac{\sin\left(\frac{m\theta}2\right)}{\frac{m\theta}2}\right)^2\left(\dfrac{\frac{n\theta}2}{\sin\left(\frac{n\theta}2\right)}\right)^2\cdot\dfrac{m^2}{n^2} …

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